, we see that in our function h{\displaystyle h} first order derivatives suffice to depict the partial differential equation. On the other hand, if h{\displaystyle h} needs no derivatives as arguments, we have due to theorem 1.4 (which you may have just proven in exercise 2) that for all continuously differentiable functions g:R→R{\displaystyle g:\mathbb {R} \to \mathbb {R} }

Since we can choose g{\displaystyle g} as a constant function with an arbitrary real value, h{\displaystyle h} does not depend on z{\displaystyle z}, since otherwise it would be nonzero somewhere for some constant function g{\displaystyle g}, as dependence on z{\displaystyle z} means that a different z{\displaystyle z} changes the value at least in one point. Therefore, the one-dimensional homogenous transport equation would be given by

h(x,t)=0{\displaystyle h(x,t)=0}

and there would be many more solutions to the initial value problem of the homogenous one-dimensional transport equation than those given by theorem and definition 1.5.

Let n∈{1,…,d}{\displaystyle n\in \{1,\ldots ,d\}} and (t,x)∈R×Rd{\displaystyle (t,x)\in \mathbb {R} \times \mathbb {R} ^{d}} be arbitrary. We choose B=[0,t]{\displaystyle B=[0,t]} and O=(−R,R){\displaystyle O=(-R,R)} for an arbitrary R>|xn|{\displaystyle R>|x_{n}|} and apply Leibniz' integral rule. We first check that all the three conditions for Leibniz' integral rule are satisfied.

1.

Since f∈C1(R×Rd){\displaystyle f\in {\mathcal {C}}^{1}(\mathbb {R} \times \mathbb {R} ^{d})}, f{\displaystyle f} is continuous in all variables (so therefore in particular in the first), and the same is true for the composition of f{\displaystyle f} with any continuous function. Thus, as all continuous functions of one variable are integrable on an interval,

exists for all y=(y1,…,yn,…,yd){\displaystyle y=(y_{1},\ldots ,y_{n},\ldots ,y_{d})} such that yn∈O{\displaystyle y_{n}\in O} (in fact, it exists everywhere, but this is the requirement of Leibniz' rule in our situation).

2.

f{\displaystyle f} was supposed to be in C1(R2){\displaystyle {\mathcal {C}}^{1}(\mathbb {R} ^{2})}, which is why

exists for all s∈B{\displaystyle s\in B} and all y=(y1,…,yn,…,yd){\displaystyle y=(y_{1},\ldots ,y_{n},\ldots ,y_{d})} such that yn∈O{\displaystyle y_{n}\in O} (in fact, it exists everywhere, but this is the requirement of Leibniz' rule in our situation).

is continuous as well as composition of continuous functions, also the function ∂xnf(s,y+v(t−s)){\displaystyle \partial _{x_{n}}f(s,y+\mathbf {v} (t-s))} is continuous in s{\displaystyle s} and yn{\displaystyle y_{n}}. Thus, due to the extreme value theorem, it is bounded for (s,yn)∈K{\displaystyle (s,y_{n})\in K}, i. e. there is a b∈R{\displaystyle b\in \mathbb {R} }, b>0{\displaystyle b>0} such that for all

for all y=(y1,…,yn,…,yd){\displaystyle y=(y_{1},\ldots ,y_{n},\ldots ,y_{d})} such that yn∈O{\displaystyle y_{n}\in O}, provided that y1,…,yn−1,yn+1,…,yd{\displaystyle y_{1},\ldots ,y_{n-1},y_{n+1},\ldots ,y_{d}} are fixed. Therefore, we might choose g(s)=b{\displaystyle g(s)=b} and obtain that